Proof that $[\Bbb{Q}(\sqrt{q_1},\dots,\sqrt{q_r}):\Bbb{Q}]=2^r$ 
Let $2\leq q_1<q_2< \dots<q_r$ be square-free natural numbers such that $mcd(q_i,q_j)=1$. Prove that the field extension $\Bbb{Q}(\sqrt{q_1},\dots,\sqrt{q_r})|\Bbb{Q}$ has degree $$[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{r}}):\Bbb{Q}]=2^r$$

I tried to prove the assertion by induction over $r$. For $r=1$, we need to prove that $\Bbb{Q}(\sqrt{q})|\Bbb{Q}$ has degree $2$, but this is clear since the minimal polynomial of $\sqrt{q}$ over $Q$ is 
$$P(t)=t^2-q$$
which has degree $2$. 
Now let's assume the assertion is true for $r=k$ and prove it for $k+1$. Consider
$$\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}})=\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}})( \sqrt{q_{k+1}})$$
Then we have the following chain of inclusions 
$$\Bbb{Q} \subset \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}}) \subset \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}})( \sqrt{q_{k+1}}) $$
Thus
$$[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}]=[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})]\cdot [\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}}):\Bbb{Q}]=2\cdot 2^k$$
since the minimal polynomial of $\sqrt{q_{k+1}}$ over $ \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{r}})$ is
$$P(t)=t^2-q_{k+1}$$
which again has degree $2$. 
That should conclude the proof. However, something seems to be wrong, since I didn't use the hypothesis that $mcd(q_i,q_j)=1$ and that each $q_i$ is square-free at all. So I'm afraid that I left something unproved, that requires the use of the hypothesis. I think that might be the part where I stated that the minimal polynomial of $\sqrt{q_{k+1}}$ over the "small field" was the one I said it was, but I really don't know how to prove it. Any help would be appreciated. Thanks in advance!
 A: I would like to give a detailed proof of why $[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})] = 2$. Then you can see where you have implicitly used what.  
Using the inductive hypothesis, $\sqrt{q_k},\sqrt{q_{k+1}},\sqrt{q_{k+1}q_{k}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$ (and because $\gcd(q_i, q_j) = 1$).
So suppose $\sqrt{q_{k+1}} \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})$.
Then there are $x,y \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$ so that $$\sqrt{q_{k+1}} = x + y \cdot \sqrt{q_{k}}$$ $$\text{(inductive hyp. provides } [\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})] = 2) \text{)}$$
(note: $x \neq 0$, because otherwise $q_{k} \mid q_{k+1}$. And also $y \neq 0$, because $\sqrt{q_{k+1}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$)
So we get $q_{k+1} = x^2 + y^2q_{k} + 2xy\sqrt{q_{k}}$ and this implies that
$$\underbrace{2xy\sqrt{q_{k}}}_{\notin\ L:= \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})} = -\underbrace{x^2}_{\in L} - \underbrace{y^2q_{k}}_{\in L} + \underbrace{q_{k+1}}_{\in \Bbb{N} \subset L} $$
which is a contradiction. So
$$\sqrt{q_{k+1}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}}) \implies [\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})] = 2$$
A: You have used it, albeit unstated. As $q_i$ is square-free, we know that $\sqrt{q_i}\notin\mathbb{Q},$ which is necessary, as otherwise it's redundant. You have also used that $\gcd(q_i,q_j)=1$ by having that $q_j\notin\mathbb{Q}(\sqrt{q_i}),$ as otherwise that might occur. With them not sharing any prime divisors, we know that the previous statement holds true.
